Just choose x = 0,1,2,3 and find y for each x, for each of the tables.
Ok, so remember that the derivitive of the position function is the velocty function and the derivitive of the velocity function is the accceleration function
x(t) is the positon function
so just take the derivitive of 3t/π +cos(t) twice
first derivitive is 3/π-sin(t)
2nd derivitive is -cos(t)
a(t)=-cos(t)
on the interval [π/2,5π/2) where does -cos(t)=1? or where does cos(t)=-1?
at t=π
so now plug that in for t in the position function to find the position at time t=π
x(π)=3(π)/π+cos(π)
x(π)=3-1
x(π)=2
so the position is 2
ok, that graph is the first derivitive of f(x)
the function f(x) is increaseing when the slope is positive
it is concave up when the 2nd derivitive of f(x) is positive
we are given f'(x), the derivitive of f(x)
we want to find where it is increasing AND where it is concave down
it is increasing when the derivitive is positive, so just find where the graph is positive (that's about from -2 to 4)
it is concave down when the second derivitive (aka derivitive of the first derivitive aka slope of the first derivitive) is negative
where is the slope negative?
from about x=0 to x=2
and that's in our range of being increasing
so the interval is (0,2)
1 and 1/2 hours each night
1+1+1+1+1= 5
1/2+1/2+1/2+1/2+1/2= 2.5
5+2.5= 7.5
7.5 hours or & and a half
hope this helps
~L
Answer:
18
Step-by-step explanation:
Each sandwich will be divided into 6 pieces. 1 sandwich equals 6 pieces, 2 sandwiches equals 12 pieces, so 3 sandwiches equals 18 pieces.
Answer:
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
Step-by-step explanation:
for
I= ∫x^n . e^ax dx
then using integration by parts we can define u and dv such that
I= ∫(x^n) . (e^ax dx) = ∫u . dv
where
u= x^n → du = n*x^(n-1) dx
dv= e^ax dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)
then we know that
I= ∫u . dv = u*v - ∫v . du + C
( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =
(u*v) - ∫v*du + C )
therefore
I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C
I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)
